Hardy-Littlewood-Polya's book Inequalities calls this result "the converse of Hölder's inequality". Cfr. Theorem 190 (161 for the infinite sum and 15 for the finite sum case). More precisely, Theorem 190 states that $(H(g)\ \text{is finite for all }g\in L^2[a, b] )\Rightarrow (f \in L^2[a, b])$, from which continuity of $H$ follows easily. This is exactly what Davide has done below.
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Giuseppe NegroApr 11 '12 at 23:49

2 Answers
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Yes, this is a consequence of the Banach-Steinhaus theorem. Without loss of generality take everything to be positive by taking absolute values everywhere and rotating the functions we evaluate the functionals on.

Consider the family of bounded functionals given by $\Lambda_n : g \mapsto \int f_n g$ where $f_n = f \chi_{f\leq n}$ (these are bounded because $f_n$ is $L^\infty$ and the measure space is finite--this last assumption is easy to remove). By Banach-Steinhaus, this family is either uniformly bounded or is infinite on a dense $G_\delta$ set. But since we take everything positive we have $|\Lambda_n(g)| \leq |H (g)| <\infty$ for all $f$. By Banach-Steinhaus there exists a uniform bound $M$ on the operator norms $\|\Lambda_n\|$ Take any $g$ with $\|g\| = 1$. Then $\int f_n g \leq M$ uniformly and so by the dominated convergence theorem $\int f g \leq M$ which gives that the map is bounded.

As a comment, a similar argument works for all $L^p$ ($1\leq p < \infty$) spaces over $\sigma-$finite measure spaces.

We work on $H:=L^2[a,b]$, which is a normed space. In fact, $f\in L^2[a,b]$, and the fact that $H$ is bounded will follow from Cauchy-Schwarz inequality. Using $\operatorname{sgn}f|g|$ for $g\in H$, we get that $fg\in L^1[a,b]$ for each $g\in L^2[a,b]$. If $f\notin H$, we can find $A_k$ pairwise disjoint such that $1\leq \int_{A_k}|f(x)|^2dx$. Put $g:=\sum_{k\geq 1}\frac 1k\chi_{A_k}\frac f{\int_{A_k}|f(x)|^2dx}$. Then $g\in H$ but $fg\notin L^1[a,b]$, a contradiction.